Changing limits of an integral The integral:
$$
\int_{-\infty}^0\frac{e^{i\alpha x}\,dx}{1+x+x^2}.
$$
If I want to change the limits of this integral so that the integral is taken from $0$ to infinity instead of minus infinity to $0$, how do I determine what change I must bring about in the integral? Also, in general if I want to change the limits of an integral how do I determine what change I must bring to the integral?
Any help would be much appreciated.
 A: Basically, if you want the limits to change from ($-\infty$ to $0$) to ($0$ to $\infty$), you should note that whenever the variable $x$ goes along the path ($0$ to $\infty$), the variable $-x$ goes along the path ($-\infty$ to $0$).
This is important because then you need to do a variable change to your original problem.  If we let $u = - x$, then our limits will change.  But the problem is, we have to worry about other changes to the integral.  Specifically, the original $dx$ will change.  If $u = - x$, then $du = -1 dx = -dx$ (do you see how I got this?).
Now that we know $u = -x$ gives us the change in limits we want, we need to solve for $x$ so that we can replace every $x$ with our new variable.  So $-u = x$.  But remember that we said we need to also worry about $du = - dx$.  Since we have a $dx$ in the integral, let's solve for this $dx$ so we can substitute.  That way, our $d$ will be in the same variable $(u)$ as the integrand.  So $-du = dx$.  Then the integral becomes:
$\int_{-\infty}^0\frac{e^{i\alpha x}}{1+x+x^2}\,dx = \int_{\infty}^{0}\frac{e^{i\alpha (-u)}}{1+(-u)+(-u)^2}\,(-du) = \int_{0}^{\infty}\frac{e^{-i\alpha u}}{1-u+u^2} \,du$
A: Hint:
With $x=g(u)$,
$$\int_a^b f(x)\,dx=\int_{g^{-1}(a)}^{g^{-1}(b)} f(g(u))\,g'(u)\,du.$$
